Chapter 5 sec5 1-5 5 - Mrs. Elmer's Classes5.1 Example 5 Using a Linear System to Solve an Application (cont.) Substitute x + 4 for y in equation (1), then solve for x. Substitute

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Systems and

Matrices

© 2008 Pearson Addison-Wesley.

All rights reserved

Sections 5.1–5.5

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-2

5.1 Systems of Linear Equations

5.2 Matrix Solution of Linear Systems

5.3 Determinant Solution of Linear

Systems

5.4 Partial Fractions

5.5 Nonlinear Systems of Equations

Systems and Matrices5


Systems of Linear Equations5.1Linear Systems ▪ Substitution Method ▪ Elimination Method ▪

Special Systems ▪ Applying Systems of Equations ▪ Solving

Linear Systems with Three Unknowns (Variables) ▪ Using

Systems of Equations to Model Data


Solve the system.

5.1 Example 1 Solving a System by Substitution (page 495)

Solve equation (2) for x: x = 1 + 2y

Replace x in equation (1) with 1 + 2y, then solve for y:

Distributive property

Replace y in equation (2) with 2, then solve for y:


The solution of the system is (5, 2). Check this

solution in both equations (1) and (2).

5.1 Example 1 Solving a System by Substitution (cont.)

Solution set: {(5, 2)}


To solve the system graphically, solve both equations

for y:

5.1 Example 1 Solving a System by Substitution (cont.)

Graph both Y1 and Y2 in the

standard window to find

that their point of

intersection is (5, 2).

2


Solve the system.

5.1 Example 2 Solving a System by Elimination (page 496)

Multiply both sides of equation (1) by 2, and then

multiply both sides of equation (2) by 3.

Add equations (3) and (4), then solve for x.


5.1 Example 2 Solving a System by Elimination (cont.)

Substitute 4 for x equation (1), then solve for y.

The solution of the system is (4, –3). Check this

solution in both equations (1) and (2).



Solution set: {(4, –3)}



The graph confirms that the solution set is {(4, –3)}.


5.1 Example 3 Solving an Inconsistent System (page 496)

Solve the system.

Multiply both sides of equation (1) by 2, then add the

resulting equation to equation (2).

The system is inconsistent.

Solution set: ø


5.1 Example 3 Solving an Inconsistent System (cont.)

The graphs of the equations are parallel and never

intersect.

3


5.1 Example 4 Solving a System with Infinitely Many Solutions (page 497)

Solve the system.

Multiply both sides of equation (2) by 3, then add the

resulting equation to equation (1).

The result indicates that the equations of the original

system are equivalent. Any ordered pair that satisfies

either equation will satisfy the system.


5.1 Example 4 Solving a System with Infinitely Many Solutions (cont.)

From equation (2), we have

The solution set (with x arbitrary) is {(x, 3x – 8)}.

From equation (2), we have

The solution set (with y arbitrary) is



The graphs of the two equations coincide.


5.1 Example 5 Using a Linear System to Solve an Application (page 498)

In the 2006 Winter Olympics in Torino, Italy, the two

countries winning the most medals were Germany and

the United States. The total number of medals won by

these two countries was 54, with Germany winning 4

more medals than the United States. How many

medals were won by each country? (Source:

www.infoplease.com)

Let x = the number of medals won by the U.S.

Let y = the number of medals won by Germany.

Then, we have the system:


5.1 Example 5 Using a Linear System to Solve an Application (cont.)

Substitute x + 4 for y in equation (1), then solve for x.

Substitute 25 for x in equation (2), then solve for y.

The U.S. won 25 medals and Germany won 29 medals.


5.1 Example 5 Using a Linear System to Solve an Application (cont.)

Verify that (25, 29) checks in the words of the original

problem.

The total number of medals won was 25 + 29 = 54.

29 is 4 more than 25.

4


5.1 Example 6 Solving a System of Three Equations with Three Variables (page 500)

Solve the system

Eliminate y by adding equations (2) and (3) to obtain

To eliminate y from another pair of equations, multiply

equation (3) by 4 and add the result to equation (1).


5.1 Example 6 Solving a System of Three Equations with Three Variables (cont.)

Solve the system consisting of equations (4) and (5) by

multiplying equation (4) by –2 and then adding the result

to equation (5) to solve for x.

Substitute –2 for x in equation (4) to solve for z.



Substitute –2 for x and –2 for z in equation (3) to solve

for y.

x = –2, y = 3, z = –4.

Verify that (–2, 3, –4) satisfies all three equations in the

original system.



Solution set:{(–2, 3, –4)}


5.1 Example 7 Solving a System of Two Equations with Three Variables (page 501)

Solve the system. Write the system with z arbitrary.

Multiply equation (1) by –2, then add the result to equation

(1) to eliminate y.

Now solve equation (3) for x.


5.1 Example 7 Solving a System of Two Equations with Three Variables (cont.)

Substitute 5z – 8 for x in equation (2), then solve for y.

With z arbitrary, the solution set is

5


5.1 Example 8 Using Curve Fitting to Find an Equation Through Three Points (page 502)

Find the equation of the parabola that

passes through the points (–5, 7), (–1, –2), and (3, 5).

The three points must satisfy the equation. Substitute

each ordered pair into the equation to obtain a system of

equations.


5.1 Example 8 Using Curve Fitting to Find an Equation Through Three Points (cont.)


(1) to eliminate c.


(3) to eliminate c.



Add equations (4) and (5), then solve for a.

Substitute ½ for a into equation (4) to solve for b.



Now substitute ½ for a and ¾ for b into equation (2) to

solve for c.

The equation of the parabola is or


5.1 Example 9 Solving an Application Using a System of Three Equations (page 503)

The table shows the number of units of protein, fat, and

fiber are in one unit of each ingredient in an animal feed.

How many units of each ingredient should be used to

make a feed that contains 35 units of protein, 38 units of

fat, and 28 units of fiber?


5.1 Example 9 Solving an Application Using a System of Three Equations (cont.)

Let x = the number of units of corn.

Let y = the number of units of soybeans.

Let z = the number of units of cottonseed.

The total amount of fat is to be 38 units, so the second

row of the table yields

Since the total amount of protein is to be 35 units, the first


Since the total amount of fiber is to be 28 units, the third


6



Multiply both sides of equation (1) by 100 and both sides

of equations (2) and (3) by 10 to obtain the equivalent

system



Multiply equation (5) by –20, then add the result to

equation (4) to eliminate y.

Multiply equation (6) by –20, then add the result to

equation (4) to eliminate y and z, then solve for x.



Substitute 60 for x in equation (7), then solve for z.

Substitute 60 for x and 20 for z in equation (6), then solve

for y.

Verify that the ordered triple (60, 40, 20) satisfies the

system formed by equations (1), (2), and (3).

The feed should contain 60 units of corn, 40 units of

soybeans, and 20 units of cottonseed.Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-34

Matrix Solution of Linear Systems5.2The Gauss-Jordan Method ▪ Special Systems


Solve the system.

5.2 Example 1 Using the Gauss-Jordan Method (page 512)

The system has the augmented matrix

Multiply each entry in the first row by ½ to get a 1 in the first row, first column position.


5.2 Example 1 Using the Gauss-Jordan Method (cont.)

Introduce 0 in the second row, first column by multiplying each element of the first row by –3 and then adding the result to the corresponding element of the second row.

Introduce 1 in the second row, second column by multiplying each element of the second row by .

7



Finally, introduce 0 in the first row, second column by multiplying each element of the second row by and then adding the result to the corresponding element of the first row.

The last matrix corresponds to the system

Verify that (–4, 5) satisfies the original system.

Solution set: {(–4, 5)}


Solve the system.

5.2 Example 2 Using the Gauss-Jordan Method (page 514)




There is already a 1 in the first row, first column.

Multiply each entry in the first row by –2, then add the result to the corresponding element of the second row to get a 0 in the second row, first column position.



Multiply each entry in the second row by , to get a 1 in the second row, second column position.

Multiply each entry in the first row by –3, then add the result to the corresponding element of the third row to get a 0 in the third row, first column position.



Multiply each entry in the second row by 4, then add the

result to the corresponding element of the third row to get

a 0 in the third row, second column position.

Multiply each entry in the third row by , to get a 1 in the

third row, third column position.



Subtract row 2 from row 1 to get a 1 in the first row,

second column position.

Multiply each entry in the third row by , then add the

result to the corresponding element of the second row to

get a 0 in the second row, third column position.

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Multiply each entry in the third row by 4, then add the

result to the corresponding element of the first row to get

a 0 in the first row, third column position.

Verify that (3, –1, –2) satisfies the original system.

Solution set: {(3, –1, –2)}


Use the Gauss-Jordan method to solve the system.

5.2 Example 3 Solving an Inconsistent System (page 515)


The first row corresponds to the equation 0x + 0y = 21, which has no solution.

Solution set: ø


Use the Gauss-Jordan method to solve the system.

Write the solution set with z arbitrary.

5.2 Example 4 Solving a System with Infinitely Many Solutions (page 516)




Solution set: {(–3z + 18, 5z – 31, z)}

It is not possible to go further. The equations that correspond to the final matrix are

Solve these equations for x and y, respectively


Determinant Solution of Linear Systems

5.3Determinants ▪ Cofactors ▪ Evaluating n × n Determinants ▪Cramer’s Rule


5.3 Example 1 Evaluating a 2 × 2 Determinant (page 523)

Graphing calculator

solution

9


5.3 Example 2 Finding Cofactors of Elements (page 525)

Find the cofactor of each of the following elements of

the matrix.

(a) 5

Since 5 is in the first row, first column, i = 1 and j = 1.

The cofactor is


5.3 Example 2 Finding Cofactors of Elements (cont.)

(b) 0

Since 0 is in the second row, first column, i = 2 and j = 1.

(c) 3

Since 3 is in the third row, third column, i = 3 and j = 3.

The cofactor is

The cofactor is



Evaluate , expanding by the third row.

It is not necessary to calculate A31 since a31 = 0.




5.3 Example 4 Applying Cramer’s Rule to a 2 × 2 System (page 528)

Use Cramer’s rule to solve the system.

By Cramer’s rule, Find D first,

since if D = 0, Cramer’s rule does not apply. If D ≠ 0,

find Dx and Dy.


5.3 Example 4 Applying Cramer’s Rule to a 2 × 2 System (cont.)

10



Graphing calculator solution

The screens support the algebraic solution.Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-56

5.3 Example 4 Applying Cramer’s Rule to a 3 × 3 System (page 529)

Use Cramer’s rule to solve the system.

Rewrite each equation in the form ax + by + cz = k.



By Cramer’s rule, Find D

first, since if D = 0, Cramer’s rule does not apply.

Use a calculator to find the determinants.

D

Dy Dz

Dx




5.3 Example 6 Showing That Cramer’s Rule Does Not Apply (page 530)

Show that Cramer’s rule does not apply to the system.

We need to show that D = 0. Expanding about column

1 gives

Thus, Cramer’s rule does not apply to the system.Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-60

Partial Fractions5.4Decomposition of Rational Expression ▪ Distinct Linear Factors ▪Repeated Linear Factors ▪ Distinct Linear and Quadratic Factors▪ Repeated Quadratic Factors

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5.4 Example 1 Finding a Partial Fraction Decomposition (page 536)

Find the partial fraction decomposition of

The degree of the numerator is greater than the

degree of the denominator, so first find the quotient.


5.4 Example 1 Finding a Partial Fraction Decomposition (cont.)

The quotient is

Now find the partial fraction decomposition for

Factor the denominator.

Partial fraction decomposition

Multiply by x(x – 1).



Let x = 0. Then equation (1) becomes

Let x = 1. Then equation (1) becomes




This is a proper fraction, and the denominator is

already factored with repeated linear factors.



Substitute 3 for x in equation (1), then solve for C.

Substitute 1 for x and 4 for C in equation (1):

We need to find a system of equations to find A and B.

Substitute –1 for x and 4 for C in equation (1):



Solve the system of equations (2) and (3) to find A

and B.

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The denominator has distinct linear and quadratic

factors, and neither is repeated.



Substitute –4 for x in equation (1), then solve for A.

Substitute 0 for x and 5 for A in equation (1):



Substitute 1 for x, 5 for A, and -2 for C in equation

(1), then solve for B.




The denominator has a linear factor and repeated

quadratic factors.



Multiply both sides by

Substitute 0 for x in equation (1):



Substitute 1 for A in equation (1), then expand and

combine terms on the right side.

To get additional equations involving the unknowns,

equate the coefficients of like powers of x on the two

sides of equation (2).

13



Substituting 0 for B in equation (3) gives D = 0.

Substituting 3 for C in equation (4) gives E = –4.


Nonlinear Systems of Equations5.5Solving Nonlinear Systems with Real Solutions ▪ Solving Nonlinear Systems with Nonreal Complex Solutions ▪ Applying Nonlinear Systems


Solve the system.

5.5 Example 1 Solving a Nonlinear System by Substitution (page 543)

Solve equation (2) for y, then substitute that expression into equation (1) and solve for x:


5.5 Example 1 Solving a Nonlinear System by Substitution (cont.)

Substituting –2 for x in equation (2) gives y = 2.

Substituting 1 for x in equation (2) gives y = 5.

Verify that the ordered pairs (–2, 2) and (1, 5) satisfy the original equations.

Solution set: {(–2, 2), (1, 5)}


5.5 Example 1 Solving a Nonlinear System by Substitution (cont.)


Solve each equation for y, and graph them in the same viewing window.

The points of intersection are (–2, 2), (1, 5).Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 5-78

Solve the system.

5.5 Example 2 Solving a Nonlinear System by Elimination (page 544)

Multiply equation (1) by –4, then add the resulting equation (3) to equation (2).

14


5.5 Example 2 Solving a Nonlinear System by Elimination (cont.)

Substitute 0 for x in equation (1), then solve for y.

Verify that the ordered pairs (0, –3) and (0, 3) satisfy the original equations.

Solution set: {(0, –3) and (0, 3)}







The points of intersection are (0, –3) and (0, 3).


Solve the system.

5.5 Example 3 Solving a Nonlinear System by a Combination of Methods (page 545)

Subtract equation (2) from equation (1) to obtain


5.5 Example 3 Solving a Nonlinear System by a Combination of Methods (cont.)

Substitute for y into equation (1) to solve for x.


5.5 Example 3 Solving a Nonlinear System by a Combination of Methods (cont.)

Solution set:

Substitute these values into equation (3) to solve for y.

15


Solve the system.

5.5 Example 4 Solving a Nonlinear System with an Absolute Value Equation (page 546)

Solving equation (2) for |x| gives |x| = y.

Since |x| ≥ 0 for all x, y ≥ 0.

Reject the negative solution.

In equation (1), the first term is x2, which is the same

as |x|2, giving


Substitute for y in equation (2) to solve for x.

5.5 Example 4 Solving a Nonlinear System with an Absolute Value Equation (cont.)

Verify that the ordered pairs and

satisfy the original system.

Solution set:







The points of intersection are and


Solve the system.

5.5 Example 5 Solving a Nonlinear System with Nonreal Complex Numbers in its Solutions (page 547)


Find the corresponding values of x by substituting the

values of y into equation (1).

5.5 Example 5 Solving a Nonlinear System with Nonreal Complex Numbers in its Solutions (cont.)

Verify that the ordered pairs

satisfy the original

system.

Solution set:

16


The length of the hypotenuse of a right triangle is

41 cm. One of the legs is 31 cm longer than the other.

Find the lengths of the two legs of the triangle.

5.5 Example 5 Using a Nonlinear System to Find the Dimensions of a Box (page 547)

Let x = the length of one leg

Let y = the length of the other leg.

Then, we have the system:


Solve equation (2) for x, then substitute that

expression into equation (1) to solve for y .

5.5 Example 5 Using a Nonlinear System to Find the Dimensions of a Box (cont.)

Reject the negative solution since length cannot be

negative.

Substitute y = 9 into equation (2) to solve for x.

The lengths of the legs are 40 cm and 9 cm.

Documents

Chapter 5 sec5 1-5 5 - Mrs. Elmer's Classes5.1 Example 5 Using a Linear System to Solve an Application (cont.) Substitute x + 4 for y in equation (1), then solve for x. Substitute